Boundary Riesz potential estimates for parabolic equations with measurable nonlinearities

1.   Introduction

In this paper, we consider parabolic equations with measurable nonlinearities and measure data

where $\mu$ is a radon measure with $|\mu|(Q^{+}_{2R}) < \infty$. Here, the parabolic half cube with the size $\rho$ is denoted as $Q^{+}_{\rho} = (0, \rho) \times (-\rho, \rho)^{n-1} \times (-\rho^{2}, 0)$ and the parabolic hyperplane with the size $\rho$ is denoted as $T_{\rho} = \{ 0 \} \times (-\rho, \rho)^{n-1} \times (-\rho^{2}, 0)$.

We will obtain the pointwise gradient estimate of $u$ in terms of Riesz potential of the right-hand side $\mu$. Here, Riesz potential of $\mu$ is defined as

For the boundary data $\psi$, we measure the pointwise oscillation of the gradient in $x'$-variable and $L^{2}$-norm of the time derivative:

For the ellipticity constants $0 < \lambda \leq \Lambda$, suppose that the nonlinearities $a(\xi, x, t)$ satisfy that

and

for any $(x, t) \in \mathbb{R}^{n+1}$, $\xi \in \mathbb{R}^{n}$ and $\zeta \in \mathbb{R}^{n}$. Also suppose that $a(\xi, x, t) = a(\xi, x^{1}, x', t)$ is Dini-continuous in $(x', t)$-variables, i.e., that there exists a function $\omega : [0, \infty) \to [0, 1]$ which is non-decreasing concave with $\lim_{\rho \searrow 0} \omega(\rho) = \omega(0) = 0$ and

satisfying that

for every $(x^{1}, x', t) \in \mathbb{R}^{n+1}$, $\xi \in \mathbb{R}^{n}$ and $y' \in \mathbb{R}^{n-1}$. Note that the nonlinearities are only merely measurable on $x^{1}$-variable.

For nonlinear parabolic equations, many authors obtained the pointwise gradient estimates by using potentials. Duzaar and Mingione considered linear growth condition in ^[1]. Kuusi and Mingione considered $p$-growth conditions and obtained Wolff potential type estimates in ^[2,3] and Riesz potential type estimates in ^[4]. Also for elliptic equations with measurable nonlinearities, the boundary pointwise gradient estimates by using Riesz potentials were obtained in ^[5], where they used the excess decay estimates of the gradient in ^[6]. In this paper, we will extend the result ^[5] to nonlinear parabolic equations with measurable nonlinearities and obtain the boundary pointwise gradient estimates by using Riesz potentials.

For the reader's further interest, we refer to ^[7] for Morrey space estimates to linear parabolic systems with measurable coefficients and measure data. We refer to ^[8] for weighted Lebesgue estimates to linear parabolic systems with measurable coefficients. Potential can be used not only for the right-hand side data of the equation but also be considered as a multiplier of the solution, see for instance ^[9], which considers existence and blow-up solution with singular potentials multiplied to the solution.

We use the following notations in this paper. Let $z$ be a typical point in $\mathbb{R}^n$, $s$ be a typical time variable and $r > 0$ be a size.

1. $z = (z^1, \cdots, z^n) = (z^1, z')$.

2. $\mathbb{R}^{n+1}_{+} = \{ (x^{1}, x', t) \in \mathbb{R}^{n+1} : x^{1} > 0 \}$, $\mathbb{R}^{n+1}_{0} = \{ (x^{1}, x', t) \in \mathbb{R}^{n+1} : x^{1} = 0 \}$.

3. $Q_{r}(z, s) = (z^{1}-r, z^{1}+r) \times \cdots \times (z^{n}-r, z^{n}+r) \times (s-r^{2}, s)$, $Q_{r} = Q_{r}(\mathbf{0}, 0)$.

4. $Q_{r}^{+}(z, s) = Q_{r}(z, s) \cap \mathbb{R}^{n+1}_{+}$, $Q_{r}^{+} = Q_{r} \cap \mathbb{R}^{n+1}_{+}$.

5. $T_{r}(z', s) = \{ 0 \} \times (z^{2}-r, z^{2}+r) \times \cdots \times (z^{n}-r, z^{n}+r) \times (s-r^{2}, s) = Q_{r}(0, z', s) \cap \mathbb{R}^{n+1}_{0}$.

6. $K_{r}(z) = (z^{1}-r, z^{1}+r) \times \cdots \times (z^{n}-r, z^{n}+r)$.

7. $K_{r}^{+}(z) = (z^{1}-r, z^{1}+r) \times \cdots \times (z^{n}-r, z^{n}+r) \cap \mathbb{R}^{n}_{+}$.

8. For $g \in L^{1}(U)$, $(g)_{U} = \rlap{-} \displaystyle {\int}_{U} g \; dx = \frac{1}{|U|} \displaystyle {\int}_{U} g \, dx$ when $|U| \not = 0$.

In view of the available approximation theory, we assume that $\mu \in L^{1}(Q^{+}_{2R})$. Without loss of generality, we shall assume that

by letting $\mu \big|_{\mathbb{R}^{n+1} \setminus Q^{+}_{2R}} = 0$. By using the concept of SOLA (Solutions Obtained by Limit of Approximations), we will remove this assumption in Corollary 1.3. Also for the boundary data, let $\psi : T_{2R} \to \mathbb{R}$ be a function such that

We obtain the following boundary pointwise gradient estimate in this paper.

Theorem 1.1. There exists a constant $c_{1} = c_{1}(n, \lambda, \Lambda) \geq 1$ such that the following holds. For some $r \in (0, R]$, assume that

If $u \in C^{0} \left(-4R^{2}, 0 \, ; \, L^{2} (K_{2R}^{+}) \right) \cap L^{2} \left(-4R^{2}, 0 \, ; \, W^{1, 1} (K_{2R}^{+}) \right)$ is a weak solution of

with the assumptions (1.2)–(1.5), then we have that

for any Lebesgue point $(x_{0}, t_{0}) = (x_{0}^{1}, x_{0}', t_{0}) \in \overline{Q_{R}^{+}}$ of $Du$ with $c = c(n, \lambda, \Lambda)$.

To deal with SOLA, we now remove the assumption (1.4).

Definition 1.2. A SOLA of (1.7) is a distributional solution $u \in L^{2} \left(-4R^{2}, 0 \, ; \, W^{1, 1} (K_{2R}^{+}) \right)$ to (1.7) such that $u$ is the limit of solutions $u_{h} \in C^{0} \left(-4R^{2}, 0 \, ; \, L^{2} (K_{2R}^{+}) \right) \cap L^{2} \left(-4R^{2}, 0 \, ; \, W^{1, 1} (K_{2R}^{+}) \right)$ to

in the sense that $u_{h} \to u$ in $L^{2} \left(-4R^{2}, 0 \, ; \, W^{1, 1} (K_{2R}^{+}) \right)$ and $L^{\infty} \owns \mu_{h} \rightharpoonup \mu$ in the sense of measures satisfying

where $\lfloor Q \rfloor$ denotes the parabolic closure of $Q$.

We finally state our main result for SOLA.

Corollary 1.3. Without (1.4), Theorem 1.1 continues to hold for SOLA of (1.7), with the estimate (1.8) for any Lebesgue point $(x_{0}, t_{0}) = (x_{0}^{1}, x_{0}', t_{0}) \in \overline{Q_{R}^{+}}$ of $Du$.

Remark 1.4. For the sake of convenience and simplicity, we employ the letter $c > 0$ and $\alpha \in (0, 1]$ throughout this paper to denote any constants which can be explicitly computed in terms of known quantities such as $n, \lambda, \Lambda$. Thus the exact values denoted by $c$ and $\alpha$ may change from line to line in each given computation.

2.   Excess decay estimates

In this section, the nonlinearities are assumed to be depending only on $\xi$ and $x^{1}$-variables with the following assumptions:

Also we assume that $a(\xi, x^{1}) : \mathbb{R}^n \times \mathbb{R} \rightarrow \mathbb{R}^n$ satisfies

for every $x^{1} \in \mathbb{R}$, $\xi \in \mathbb{R}^n$, $\zeta \in \mathbb{R}^n$ and for some constants $0 < \lambda\leq \Lambda$, $\Gamma\geq 0$.

We obtain the boundary excess-decay estimates for parabolic equations in this section. Under the assumptions (2.1), (2.2) and

let $g$ be a weak solution of

where $\gamma' = (\gamma^{2}, \cdots, \gamma^{n}) \in \mathbb{R}^{n-1}$ and $(x_{0}, t_{0}) = (x_{0}^{1}, x_{0}^{2}, \cdots, x_{0}^{n}) = (x_{0}^{1}, x_{0}', t_{0})$.

Remark 2.1. If the nonlinearity $a(\xi, x^{1})$ is only defined on $0 < x^{1} < 3r$, then one can easily extend $a(\xi, x^{1})$ to satisfy (2.1) and (2.2), which does not effect the results in this paper.

The nonlinearity $a(\xi, x^{1})$ only depends on $x^{1}$-variable and $g = \gamma' \cdot x'$ on $T_{3r}(x_{0}', t_{0})$. So one can use the difference quotient method to find that $D_{k}g - \gamma^{k}$ $(k \in \{2, \, 3, \, \cdots, \, n \})$ is weakly differentiable in $Q_{2r}^{+}(x_{0}, t_{0})$. Moreover, one can show that $D_{k}g - \gamma^{k} \in W^{1, 2}(Q_{2r}^{+}(x_{0}, t_{0}))$. By differentiating (2.4) with respect to $x^{k}$-variable, one can get that

where $a_{ij}(x, t) = \frac{\partial a^{i}}{\partial \xi_{j}} (Dg, x^{1})$ satisfies that

for any $(x, t) \in Q_{2r}^{+}(x_{0}, t_{0})$ and $\zeta \in \mathbb{R}^{n}$ with $1\leq i, j\leq n$.

To obtain boundary estimates, we next extend the equation (2.5) from $Q_{2r}^{+}(x_{0}, t_{0})$ to $Q_{2r}(x_{0}, t_{0})$. For $k \in \{2, \, 3, \, \cdots, \, n \}$, we let

Then

Let $\hat{a}_{ij}(x, t)$ be an extension of $a_{ij}(x, t)$ from $Q_{2r}^{+}(x_{0}, t_{0})$ to $Q_{2r}(x_{0}, t_{0})$ defined as

for $(x^1, x', t) \in Q_{2r}(x_{0}, t_{0})\setminus Q_{2r}^{+}(x_{0}, t_{0})$. Then one can check from (2.6) that

for any $(x, t) \in Q_{2r}(x_{0}, t_{0})$ and $\zeta \in \mathbb{R}^{n}$. Then we obtain from (2.8) and (2.9) that $g_{k}$ is a weak solution of the parabolic equation

From [10, Chapter 6], we have an excess decay estimate for linear parabolic equations, which can be applied to (2.11).

Lemma 2.2. Under the assumptions

let $w$ be a weak solution of

Then we have that

where $c = c(n, \lambda, \Lambda)$ and $\alpha = \alpha(n, \lambda, \Lambda)\in(0, 1]$.

With Lemma 2.2 and the energy estimate, we obtain the following lemma.

Lemma 2.3. Suppose that $k \in \{ 2, 3, \cdots, n \}$. For $g_{k}$ in (2.7), we have that

and

for any $Q_{\rho}(y, s) \subset Q_{2r}(x_{0}, t_{0})$ and $0 < \tau < \rho$.

Proof. Let $k \in \{ 2, 3, \cdots, n \}$ be an arbitrary integer. The estimate (2.12) follows by applying Lemma 2.2 and (2.10) to (2.11).

Next, we choose a cut-off function $\eta \in C_{c}^{\infty}(Q_{\rho}(y, s))$ with

We have from (2.11) that

Since $\eta \in C_{c}^{\infty}(Q_{\rho}(y, s))$ and $Q_{\rho}(y, s) \subset Q_{2r}(x_{0}, t_{0})$, we test the above equation by $\big[ g_{k} - \zeta^{k} \big] \eta^{2}$ to find that

Since $\eta \in C_{c}^{\infty}(Q_{\rho}(y, s))$, one can check that $\displaystyle {\int}_{Q_{\rho}(y, s)} \partial_{t} \left\{\left(\big[ g_{k} - \zeta^{k} \big] \eta\right)^{2} \right\} \, dx dt \geq 0$. So by (2.10),

By Young's inequality,

So (2.13) follows from (2.14).

Now, we extend $g_{t}$ from $Q_{2r}^{+}(x_{0}, t_{0})$ to $Q_{2r}(x_{0}, t_{0})$ and obtain some estimates on the extended function of $g_{t}$. Since $g$ is a weak solution of (2.4), one can use the difference quotient method to show that $g_{t} = \partial_{t} g$ is a weak solution of

In view of (2.4), one can check that $g_{t} = 0$ on $T_{2r}(x'_{0}, t_{0})$. So

defined as

So we find from (2.9) and (2.16) that

Then we have the following energy estimate for $g_{n+1}$ in (2.18).

Lemma 2.4. If $g_{n+1}$ is a weak solution of (2.18), then we have that

for any $Q_{\rho}(y, s) \subset Q_{2r}(x_{0}, t_{0})$.

Proof. Choose a cut-off function $\eta \in C_{c}^\infty (Q_{\rho}(y, s))$ with

We test (2.18) by $\varphi = \eta^2 g_{n+1}$ to find that

Then a direct calculation gives that

From the fact that $\eta \in C_{c}^{\infty}(Q_{\rho}(y, s))$, we get

By combining the above two equalities and applying the elliptic condition (2.6), we get

So the lemma follows from the choice of the cut-off function $\eta$ in (2.19).

Since the nonlinearity $a(\xi, x^{1})$ depends on $x^{1}$-variable, we obtain an excess decay estimate in terms of $a^{1}(Dg, x^{1})$ instead of $D_{1}g$. Let $g_{1}$ be the even extension of $a^{1}(Dg, x^{1})$ from $Q_{2r}^{+}(x_{0}, t_{0})$ to $Q_{2r}(x_{0}, t_{0})$ defined as

So from (2.7), (2.17) and (2.21), we have following extensions from $Q_{2r}^{+}(x_{0}, t_{0})$ to $Q_{2r}(x_{0}, t_{0})$:

Then we define $G : Q_{2r}(x_{0}, t_{0})\rightarrow \mathbb{R}^{n}$ as

The desired excess-decay estimate will be obtained with the function $G$ in (2.23).

With Lemma 2.4, we estimate $g_{n+1}$ by using the function $G$ in (2.23).

Lemma 2.5. For $g_{n+1}$ in (2.17), we have that

for any $Q_{\rho}(y, s) \subset Q_{2r}(x_{0}, t_{0})$ and $\zeta = \left(\zeta^{1}, \zeta^{2}, \cdots, \zeta^{n} \right) \in \mathbb{R}^{n}$.

Proof. Let $d_{0} = \tau$ and $d_{\infty} = \rho$. Let

Choose a cut-off function $\eta \in C_{c}^\infty \left(Q_{e_{m}}(x_{0}, t_{0})\right)$ with

Since $g_{t} = 0$ on $T_{2r}(x_{0}', t_{0})$, we test (2.4) by $\eta^2 g_{t}$ to find that

Since $\eta \in C_{c}^\infty \left(Q_{e_{m}}(x_{0}, t_{0})\right)$ and $g_{t} = 0$ on $T_{2r}(x_{0}', t_{0})$, one can easily check that

Then for $\kappa > 0$, Young's inequality implies that

By using integration by parts, for any $k \in \{ 2, 3, \cdots, n \}$ we have that

Then Young's inequality implies that

By (2.7), $g_{k}$ is the odd extension of $D_{k}g - \gamma^{k}$ from $Q_{2r}^{+}(x_{0}, t_{0})$ to $Q_{2r}(x_{0}, t_{0})$, which implies

By combining the above two estimates, we apply $\tau = e_{m}$ and $\rho = d_{m+1}$ in Lemma 2.3. Then

because $d_{m+1} - e_{m+1} = \frac{d_{m+1} - d_{m}}{2} = \frac{\rho - \tau}{2^{m+1}}$. So we obtain from (2.25), (2.26) and (2.27) that

By (2.22), $g_{1}$ is the even extension of $a^{1} \left(Dg, x^{1} \right)$, $g_{k}$ $(k \in \{ 2, 3, \cdots, n\})$ is the odd extension of $D_{k}g - \gamma^{k}$ and $g_{n+1}$ is the odd extension of $g_{t}$ from $Q_{2r}^{+}(x_{0}, t_{0})$ to $Q_{2r}(x_{0}, t_{0})$. Thus

Now, take $\rho = e_m$ and $r = d_{m+1}$ in Lemma 2.4 to find that

Take $\kappa$ so that $\frac{c_{1} \kappa 4^{m}}{(r-\rho)^2} = \frac{1}{48}$. By combining (2.24), (2.28) and (2.29), we have

Multiply (2.30) by $\frac{1}{8^m}$ and sum it from $m = 0$ to $\infty$. Then we have that

Thus from (2.31) and the fact that $d_{0} = \tau$, we have

which finishes the proof.

To obtain the desired excess-decay estimate on $G = \left(g_{1}, \cdots, g_{n} \right)$, we will use Poincaré's inequality. In Lemma 2.3 and Lemma 2.4, the derivatives $Dg_{2}, \cdots, Dg_{n}$ and $Dg_{n+1}$ were obtained. So it only remains to obtain the following estimate on $Dg_{1}$.

Lemma 2.6. For $g_{1}$ in (2.22), $Dg_{1} \in L^{2}(Q_{r}(x_{0}, t_{0}))$ exists with the estimate

Proof. We discover from Lemma 2.3 and Lemma 2.5 that

for any $k \in \{ 2, 3, \cdots, n \}$. It follows from (2.22) that

Since $g$ is a weak solution of (2.4) and the nonlinearities $a(\xi, x^{1})$ are independent of $x^{k}$-variable for any $k \in \{ 2, 3, \cdots, n \}$, we have from (2.32) that

From (2.6), $a_{ij}(x, t)$ is uniformly elliptic. So we find from (2.22) that

By (2.22), $g_{1}$ is the even extension of $a^{1} (Dg, x^{1})$, $g_{k}$ $(k \in \{ 2, 3, \cdots, n\})$ is the odd extension of $D_{k}g - \gamma^{k}$ and $g_{n+1}$ is the odd extension of $g_{t}$ from $Q_{2r}^{+}(x_{0}, t_{0})$ to $Q_{2r}(x_{0}, t_{0})$. So the lemma follows by extending the above estimate from $Q^{+}_{r}(x_{0}, t_{0})$ to $Q_{r}(x_{0}, t_{0})$.

We obtain the following excess-decay estimate and $L^{\infty}$-estimate of $g_{n+1}$.

Lemma 2.7. For the odd extension $g_{n+1}$ of $g_{t}$ in (2.22), we have that

and

for any $Q_{\rho}(y, s) \subset Q_{r}(x_{0}, t_{0})$ and $0 < \tau \leq \rho$ where $\alpha = \alpha(n, \lambda, \Lambda) \in (0, 1]$.

Proof. From (2.18), $g_{n+1}$ is a weak solution of

By using (2.6) and applying Lemma 2.2 to (2.34), we find that

for any $Q_{\rho}(y, s) \subset Q_{r}(x_{0}, t_{0})$ and $0 < \tau \leq \rho$. The $L^{\infty}$-estimate of $g_{n+1}$ in (2.33) follows by applying Campanato type embedding to the excess-decay estimate (2.35).

Recall the definition of $G$ in (2.22) and (2.23). In view of Lemma 2.4 and Lemma 2.6, one can estimate $DG$ and $\partial_{t}G$ as

which implies that

for any $Q_{\rho}(y, s) \subset Q_{r}(x_{0}, t_{0})$. Here, $Dg_{k}$ $(k \in \{2, 3, \cdots, n \})$, $g_{n+1}$ and $Dg_{n+1}$ were estimated in Lemma 2.3, Lemma 2.5 and Lemma 2.4 respectively. So we use Sobolev type embeddings to have the following reverse Hölder type inequality.

Lemma 2.8. For $G$ in (2.22) and (2.23), we have that

for any $Q_{\rho}(y, s) \subset Q_{r}(x_{0}, t_{0})$. Here, $2^{*} = \frac{2(n+1)}{n-1} > 2$ is the Sobolev conjugate for $(n+1)$-dimension.

Proof. Fix any $Q_{\rho}(y, s) \subset Q_{r}(x_{0}, t_{0})$. Choose arbitrary constants $\frac{\rho}{2} \leq \tau_{1} < \tau_{2} \leq \rho$. Then by the Sobolev type embedding,

Here, the Sobolev conjugate for $(n+1)$-dimension is denoted as $2^{*} = \frac{2(n+1)}{n-1} > 2$. We have from (2.37) that

Since $\frac{\rho}{2} \leq \tau_{1} < \tau_{2} \leq \rho$, we have from (2.13) in Lemma 2.3 and Lemma 2.5 that

Since $\frac{\rho}{2} \leq \tau_{1} < \tau_{2} \leq \rho$, we have from Lemma 2.4 and Lemma 2.5 that

Since $\frac{\tau_{1}^{2}}{\left(\tau_{2}-\tau_{1}\right)^{2}} \geq 1$, by combining the above four estimates, we get

By the interpolation inequality, we get

Since $2^{*} = \frac{2(n+1)}{n-1} > 2$, we obtain from Young's inequality that

Since $\frac{\rho}{2} \leq \tau_{1} < \tau_{2} \leq \rho$ were chosen arbitrarily, by [11, Lemma 4.3], we get

and the lemma follows.

By using (2.37), we apply Poincaré's inequality to obtain the desired excess-decay estimate on $G = \left(g_{1}, \cdots, g_{n} \right)$. We remark that $Dg_{1}$, $Dg_{k}$ $(k \in \{2, \cdots, n\})$, $Dg_{n+1}$ and $g_{n+1}$ were estimated in Lemma 2.6, Lemma 2.3, Lemma 2.4 and Lemma 2.5 respectively.

Lemma 2.9. For $G$ in (2.22) and (2.23), we have that

for any $Q_{\rho}(y, s) \subset Q_{r}(x_{0}, t_{0})$ where $\alpha = \alpha(n, \lambda, \Lambda) \in (0, 1]$.

Proof. Assume that $8\tau \leq \rho$, otherwise the lemma holds from the dilation. We claim that

From Poincaré's inequality and (2.37), we have that

By taking $\zeta^{k} = \left(g_{k} \right)_{Q_{2\tau}(y, s)}$ in Lemma 2.3, we get

By the assumption $8 \tau \leq \rho$, we have from (2.12) in Lemma 2.3 that

By combining the above two estimates, we get

By the assumption $8 \tau \leq \rho$, we have from (2.33) in Lemma 2.7 that

Also we take $\zeta = \left(G \right)_{Q_{\frac{\rho}{2}}(y, s)}$ in Lemma 2.5 to find that

By combining the above two estimates, we get

The claim (2.38) holds from (2.39), (2.40) and (2.41). With Hölder's inequality, the lemma follows from (2.38) and Lemma 2.8 by taking $\zeta = \left(G \right)_{Q_{\rho}(y, s)}$.

3.   Comparison estimates

For the comparison estimates on $Q_{r}(x_{0}, t_{0})$, we handle the interior case $x_{0}^{1} > r$ in Subsection 3.1 and the boundary case $0 \leq x_{0}^{1} \leq r$ in Subsection 3.2. From [12, Lemma 4.1], the absolute value of measurable nonlinearities $|a(\xi, x^{1})|$ is comparable to $|\xi|$. In fact, one can easily modify the proof of [12, Lemma 4.1] to obtain the following result, where the nonlinearities depend on $\xi$, $x$ and $t$.

Lemma 3.1. Suppose that (1.2). For any $(x, t) \in \mathbb{R}^{n+1}$, we have that

3.1.   Interior comparison estimates

For a weak solution $u$ of

let $v$ and $g$ be the weak solution of

and

where $\partial_{p}$ denotes the parabolic boundary. By repeating the proof of the comparison estimate for $Du$ and $Dv$ such as in [1, Lemma 4.1] and [3, Lemma 4.1], one can prove that

By repeating the proof such as in [1, Lemma 4.2], one can prove that

So we obtain that

We set

From [13, Lemma 4.9], we have that

From Lemma 3.1, we have that $|Du| \leq c|U|$. Since $|U-G| \leq c|Du-Dg|$, we find from (3.1) that

So we obtain from (3.3) that

One can easily check that

so that

for any $0 < 2\rho \leq r$.

3.2.   Comparison estimates near the boundary

To handle the boundary case, we assume that

For the boundary data, we assume that

We regard the boundary data $\psi$ as a function in $Q^{+}_{4r}(x_{0}, t_{0})$ by defining $\psi(x^1, x', t) = \psi(0, x', t)$ for every $0 < x^1 < 4r$. For a weak solution $u$ of

let $v$, $w$ and $g$ be the weak solution of

and

We have from (3.6) and (3.7) that $v = u = \psi$ on $T_{4r}(x_{0}', t_{0})$. So from (3.7) and (3.8), we have that

By following the proof of [3, Lemma 4.1] (although ^[3] considered $p$-Laplace type equations), we obtain the comparison estimate for $Du$ and $Dv$ to our problems. The proof for Lemma 3.2 is similar to that of [3, Lemma 4.1], but we give the proof for the convenience of the readers.

Lemma 3.2. Under the assumption (3.5), we have that

Proof. We first claim that

To prove the claim (3.11), fix $\tau \in (t_{0}-r^{2}, t_{0})$. For $m \geq 1$, let $\phi : \mathbb{R} \to [0, 1]$ be a function only depending on $t$-variable defined as

Here, we remark that we will let $m \to \infty$ later. We also define

which implies that

Now, we test (3.6) and (3.7) by $\eta_{1, \epsilon}$ to find that

One can check that

Since $\phi = 0$ on $K_{4r} \times \left\{ t_{0} \right\}$ and $u = v$ on $K_{4r} \times \left\{ t_{0}-r^{2} \right\}$, integration by parts gives

We obtain from (3.13) and (3.14) that

By Lebesgue dominated convergence theorem, we get

On the other hand, by ellipticity condition (1.2), we have that

Since $0 \leq \eta_{1, \epsilon} \leq 1$, we find that

So we obtain from (3.16) that

By letting $m \to \infty$ for $\phi$, we find that

Since $\tau \in (t_{0}-r^{2}, t_{0})$ was chosen arbitrarily, this proves the claim (3.11).

Since $\phi$ is non-increasing, we have that $\partial_{t} \left\{ - \phi(t) \right\} \geq 0$, which implies that

So it follows from (3.16) that

Since $0 \leq \eta_{1, \epsilon} \leq 1$, we also obtain from (3.13) that

We next claim that

for $\beta > 0$ and $\nu > 1$. To this end, for $\epsilon > 0$, we test (3.6) and (3.7) by

which implies that

By the same reasoning for (3.15), we get

Since $\phi$ is non-increasing, we have that $\partial_{t} \left\{ - \phi(t) \right\} \geq 0$. So the above equality and (3.11) give that

One can compute that

Here, we have from (3.17) that

and

With the above four estimates, we find from (3.20) that

By the definition of $\eta_{1, \epsilon}$, one can see that

which implies that

By letting $\tau \to t_{0}$ and $m \to \infty$, the claim (3.18) follows from (3.12) and (3.21).

Choose $\beta = \left(\rlap{-} \displaystyle {\int}_{Q^{+}_{4r}(x_{0}, t_{0})} |u-v|^{\frac{n+1}{n}} \, dx dt \right)^{\frac{n}{n+1}}$ and $\nu = \frac{n+1}{n}$. Then by the paraoblic Sobolev embedding (see for instance [14, Chapter 1, Proposition 3.1]), we get

It follows from (3.11) that

By Hölder's inequaltiy, (3.17) and (3.22), we obtain that

Since $\left| Q^{+}_{4r}(x_{0}, t_{0}) \right| \geq c \, r^{n+2}$, the lemma follows.

We also prove the comparison estimate between $Dv$ and $Dw$ as follows.

Lemma 3.3. Under the assumption (3.5), we have that

Proof. With $v - w + \psi - D_{x'}\psi(x_{0}', t_{0}) \cdot x'$, test (3.7) and (3.8). Fix $\tau \in (t_{0}-16r^{2}, t_{0})$. Then

Recall that $\psi(x^{1}, x', t) = \psi(x', t)$. It follows from (1.2) and Young's inequality that

By a direct calculation,

From Young's inequality, we get that

Thus we find that

where

Since $\tau \in (t_{0}-16r^{2}, t_{0})$ was arbitrary chosen, we find that

which proves the lemma.

We use the following reverse Hölder type inequality for comparing $Dw$ and $Dg$.

Lemma 3.4. Under the assumption (3.5), we have that

Proof. We let

We obtain from (3.5), (3.6) and (3.7) that

It follows from (3.10) and (3.23) that

Define $\hat{w}$ as the zero extension of $w - \gamma' \cdot x'$ from $Q^{+}_{4r}(x_{0}, t_{0})$ to $Q_{4r}(x_{0}, t_{0})$. Then

Let $2_* = \frac{2n}{n+2}$. Then by dividing into two cases (1) $Q_{2\rho}(y, s) \subset \mathbb{R}^{n+1}_{+}$ and (2) $Q_{2\rho}(y, s) \not \subset \mathbb{R}^{n+1}_{+}$, we prove the following assertion that

for any $Q_{3\rho}(y, s) \subset \subset Q_{4r}(x_{0}, t_{0})$.

Choose $Q_{2\rho}(y, s) \subset \subset Q_{4r}(x_{0}, t_{0})$. First, suppose that $Q_{2\rho}(y, s) \subset \mathbb{R}^{n+1}_{+}$. Then $Q_{2\rho}(y, s) \subset Q^{+}_{4r}(x_{0}, t_{0})$. Fix $\rho \leq r_{1} < r_{2} \leq 2\rho$. Let $\eta \in C_{c}^{\infty}(Q_{{r_{2}}}(y, s))$ be a cut-off function with

Fix $\tau \in \left(s-r_{1}^{2}, s \right)$. Take a test function $\left\{ w-(w)_{Q_{r_{1}}(y, s)} \right\} \eta^2$ for (3.8) to find that

One can check that

which implies that

In view of (3.27) and (3.28), we apply the ellipticity condition (1.2) to find that

First, apply Young's inequality. Then (3.26) gives that

where we used that

which holds from that $\tau \in \left(s-r_{1}^{2}, s \right)$ and $\rho \leq r_{1} < r_{2} \leq 2\rho$.

Since $\tau \in \left(s-r_{1}^{2}, s \right)$ was arbitrary chosen, we find from (3.28) and (3.29) that

for any $\rho \leq r_{1} < r_{2} \leq 2\rho$. By the parabolic Sobolev embedding (see for instance [14, Chapter 1,Proposition 3.1]), we get

So one can use Young's inequality to find that

Let $g(\theta) = \displaystyle {\int}_{Q_{\theta}(y, s)} |Dw|^{2} \, dx dt + \sup\limits_{ \tau \in \left(s-\theta, s \right)} \displaystyle {\int}_{K_{\theta}(y)} \left[ w(x, \tau) -(w)_{Q_{\theta}(y, s)} \right]^{2} \, dx$. Then

Since $\rho \leq r_{1} < r_{2} \leq 2\rho$ were arbitrary chosen, we obtain from [11, Lemma 4.3] that

which implies that

Thus

Recall from (3.24) that $\hat{w} = w - \gamma' \cdot x'$ in $Q_{2\rho}(y, s)$, which implies that

So the assertion (3.25) holds for the case $Q_{2\rho}(y, s) \subset \mathbb{R}^{n+1}_{+}$.

Now, suppose that $Q_{2\rho}(y, s) \not \subset \mathbb{R}^{n+1}_{+}$. Then by the fact that $Q_{3\rho}(y, s) \subset Q_{4r}(x_{0}, t_{0})$,

Fix $\rho \leq r_{1} < r_{2} \leq 2\rho$. Let $\eta \in C_{c}^{\infty}(Q_{{r_{2}}}(y, s))$ be a cut-off function with

Since $Q_{3\rho}(y, s) \subset Q_{4r}(x_{0}, t_{0})$, it follows from (3.31) that $\eta \in C_{c}^{\infty}(Q_{4r}(x_{0}, t_{0}))$. We also have from (3.24) that $\hat{w} = 0$ in $Q_{4r}(x_{0}, t_{0})\setminus \mathbb{R}^{n+1}_{+}$. So we discover that

Fix $\tau \in \left(s-r_{1}^{2}, s \right)$. Take the test function $\hat{w}\eta^2 \in W^{1, 2}_{0} \left(Q^{+}_{4r}(x_{0}, t_{0}) \right)$ for (3.8) to find that

where we used (3.31) and that $Q_{3\rho}(y, s) \subset Q_{4r}(x_{0}, t_{0})$. By a direct calculation,

From (3.24), we have that $D\hat{w} = Dw - \gamma$ in $\mathbb{R}^{n+1}_{+}$. So (1.2) gives that

Since $\eta \in C_{c}^{\infty}(Q_{{r_{2}}}(y, s))$, we have from Young's inequality that

By (3.24), $\hat{w} = 0$ in $Q_{4r}(x_{0}, t_{0})\setminus \mathbb{R}^{n+1}_{+}$. Since $Q_{2\rho}(y, s) \subset Q_{4r}(x_{0}, t_{0})$, we have that

and it follows from (3.32) that

Since $\tau \in \left(s-r_{1}^{2}, s \right)$ was arbitrary chosen, we have from (3.31) that

From (3.30) and that $\hat{w} = 0$ in $Q_{4r}(x_{0}, t_{0})\setminus \mathbb{R}^{n+1}_{+}$, we have the Sobolev-Poincaré type inequality in [15, Theorem 3.16] to get

So with Young's inequality, one can use the above two inequalities to find that

Let $g(\theta) = \sup\limits_{ \tau \in \left(s-\theta, s \right)} \displaystyle {\int}_{K_{\theta}(y)} \left[ \hat{w}(x, \tau) \right]^{2} \, dx + \displaystyle {\int}_{Q_{\theta}(y, s)} |D\hat{w}|^{2} \, dx dt$. Then

Since $\rho \leq r_{1} < r_{2} \leq 2\rho$ were arbitrary chosen, we obtain from [11, Lemma 4.3] that

which implies that

So (3.25) holds for the case $Q_{2\rho}(y, s) \not \subset \mathbb{R}^{n+1}_{+}$.

By dividing into two cases (1) $Q_{2\rho}(y, s) \subset \mathbb{R}^{n+1}_{+}$ and (2) $Q_{2\rho}(y, s) \not \subset \mathbb{R}^{n+1}_{+}$, we have the assertion (3.25). Since $Q_{3\rho}(y, s) \subset \subset Q_{4r}(x_{0}, t_{0})$ in (3.25) was arbitrary chosen, by applying [16, Lemma 3.1] for $s = |\gamma|$ and $\chi_{0} = \frac{2}{2_*} > 1$ (with a suitable covering argument because the size is $3\rho$ in the right-hand side of (3.25) not $2\rho$), we have that

Since $\gamma = (0, D_{x'}\psi(x_{0}', t_{0}))$, the lemma follows from (3.5) and (3.24).

In Lemma 3.4, we obtained the reverse Hölder type inequality. So we can obtain the following comparison estimate for $Dw$ and $Dg$.

Lemma 3.5. Under the assumption (3.5), we have that

Proof. By using $w-g$, test (3.8) and (3.9). Then we get

We obtain from (1.2) and (1.3) that

With Young's inequality and Lemma 3.4, we get that

From Hölder's inequality, the lemma follows.

With Lemma 3.2, Lemma 3.3 and Lemma 3.5, the comparison estimates for $Du$ and $Dg$ will be obtained. We now have the comparison estimate Lemma 3.6 and the excess decay estimate Lemma 2.9, so the remaining proof is similar to the elliptic case in ^[5]. However, for the sake of the completeness, we give a detailed proof.

Lemma 3.6. Under the assumption (3.5), we have that

Proof. By Lemma 3.5,

In view of Lemma 3.2 and Lemma 3.3,

From the above two estimates, the lemma follows.

Recall the definition of $G$ in (3.2), which is an extension of $(a^{1}(Dg, x^{1}, x_{0}', t_{0}), D_{2}g, \cdots, D_{n}g)$. In Lemma 3.6, we obtained an excess decay estimate of $G$. For an extension $U : Q_{4r}(x_{0}, t_{0})\to \mathbb{R}^{n+1}$ of $(a^{1}(Du, x^{1}, x_{0}', t_{0}), D_{2}u - D_{2}\psi, \cdots, D_{n}u - D_{n}\psi)$ which is defined as

where

we derive an excess decay estimate in the following lemma.

Lemma 3.7. Under the assumption (3.5), we have that

for any $0 < 2\rho \leq r$.

Proof. We set $G : Q_{3r}(x_{0}, t_{0})\to \mathbb{R}^{n}$ as

where

We have from (3.5) and Lemma 2.9 that

for any $0 < 2\rho \leq r$. In view of (3.36) and (3.37), we discover that

for any $k \in \{2, 3, \cdots, n\}$. Since $0 \leq x_{0}^{1} \leq r$, we have from (3.36) and (3.37) that

for any $k \in \{2, 3, \cdots, n\}$. From (3.36) and (3.37), we find that

which implies that

because $0 \leq x_{0}^{1} \leq r$ in (3.40). For $U$ and $G$ in (3.35) and (3.37), we have from (3.39) and (3.40) that

On the other hand, Lemma 3.6 gives that

We find from Lemma 3.1 that $|Du| \leq c|U|$. So by combining the above two estimates,

From (3.38) and (3.41), we have that

for any $0 < 2\rho \leq r$. One can easily check that

and the lemma follows.

4.   Pointwise Riesz potential estimates

The remaining proof is similar to the elliptic case ^[5], but we give a detailed proof for the completeness. For a weak solution $u$ of (1.7), define $U : Q_{2R} \to \mathbb{R}^{n}$ as

where

Lemma 4.1. For any $(x_{0}, t_{0}) \in \overline{Q_{R}^{+}}$ and $0 < \rho \leq 4r \leq R$, we have that

Proof. If $r < 2\rho \leq 8r$, then one can directly check that

and the lemma follows.

We now suppose that $0 < 2\rho \leq r$. Assume that $x_{0}^1 > r$, which is the interior case. Then $Q_{r}(x_{0}, t_{0})\subset \mathbb{R}^{n+1}_{+}$. The definition of $U$ in (3.2) and (4.1) are different, but the value of $U - (U)_{Q_{\rho}(x_{0}, t_{0})}$ in $Q_{\rho}(x_{0}, t_{0})$ and $U - (U)_{Q_{r}(x_{0}, t_{0})}$ in $Q_{r}(x_{0}, t_{0})$ from (3.2) and (4.1) differs by at most $\mathop {{\text{osc}}}\limits_{T_{4r}(x_{0}', t_{0})} D_{x'}\psi$. Also the value of $U$ from (3.2) and (4.1) differs by at most $|D_{x'}\psi| \leq |D_{x'}\psi(x_{0}', t_{0})| + \mathop {{\text{osc}}}\limits_{T_{4r}(x_{0}', t_{0})} D_{x'}\psi$ in $Q_{4r}(x_{0}, t_{0})$. So we find from (3.4) that

So the lemma holds when $x_{0}^{1} > r$.

On the other hand, if $0 \leq x_{0}^{1} \leq r$, Lemma 3.7 implies that

where $U$ in Lemma 3.7 was defined in (3.35) which is same to that in (4.1). So the lemma holds when $0 \leq x_{0}^{1} \leq r$.

Now, we prove Theorem 1.1.

Proof of Theorem 1.1. Let $x_{0} = (x_{0}^{1}, x_{0}', t_{0}) \in \overline{Q_{R}^{+}}$ be a Lebesgue point of $Du$. From Lemma 4.1, we get that

for any $0 < \rho_{1} \leq \rho_{2} \leq R$. Choose $\delta = \delta(n, \lambda, \Lambda) \in (0, 1/2]$ satisfying that

which implies that

for any $0 < \rho \leq R$. We choose the constant $c_{1} = c_{1}(n, \lambda, \Lambda) \geq 1$ in (1.6) as

Then from the assumption (1.6) in Theorem 1.1, one can check that

For $i = 1, 2, \cdots$, let $r_{i} = \delta^{i}r$, $Q_{i} = Q_{r_{i}}(x_{0}, t_{0})$, $Q_{i}^{+} = Q_{r_{i}}^{+}(x_{0}, t_{0})$, $T_{r_{i}} = T_{r_{i}}(x_{0}', t_{0})$, $E_{i} = \rlap{-} \displaystyle {\int}_{Q_{i}} |U - (U)_{Q_{i}}| \, dx dt$, $F_{i} = \left| \rlap{-} \displaystyle {\int}_{Q_{i}} U \, dx dt \right|$ and

Choose $\rho = r_{i}$ in (4.4). Then we get that

Here, we obtain that

because of that

Since $\delta = \delta(n, \lambda, \Lambda) \in (0, 1/2]$, one can directly check that

From (4.5), we get

which implies that $c_{3} \omega(r_{i}) \leq 1/16$ for $i = 0, 1, 2, \cdots$. So from (4.7), we have that

Sum the above inequality over $j \in \{ 0, 1, \cdots, i-1 \}$. Then we get

To simplify the computation, define $\nu = 4 \delta^{-n} c_{3} \sum\limits_{j = 0}^{\infty} \nu_{j}$. Then we have from (4.6) and (4.8) that

We claim that

To this end, since $\delta \in (0, 1/2]$ and $F_{0} = \left| \rlap{-} \displaystyle {\int}_{Q_{0}} U \, dx dt \right|$, the claim follows holds when $i = 0$. To use induction, we next assume that (4.10) holds when $0, 1, \cdots, i$, which means that

By a direct computation,

which implies that

By using (4.9), we obtain that

We discover from (4.8) and (4.11) that

which implies that

Since $F_{0} = \left| \rlap{-} \displaystyle {\int}_{Q_{0}} U \, dx dt \right|$, we have that $F_{0} \leq \rlap{-} \displaystyle {\int}_{Q_{0}} |U| \, dx dt$. Also we have that

Thus

and the claim (4.10) holds when $i+1$. So by an induction, the claim (4.10) holds for $i = 0, 1, 2, \cdots$.

We have from (4.9) that

which implies that

By (4.8) and (4.10), we get that

So we find from (4.10) that

In view of (4.12), we get that

By the definition $Q_{i} = Q_{\delta^{i}r} (x_{0}, t_{0})$,

With the estimate (4.13), it is ready to estimate $Du$. Recall from (4.1) and Lemma 3.1 that

Since $x_{0} \in \overline{Q_{R}^{+}}$, we find from (4.14) that

Since $x_{0} \in \overline{Q_{R}^{+}}$ and $U$ is an extension defined in (4.2), we have from (4.14) that

By using the above two estimates, we have from (4.13) that

where we used that $\sup\limits_{T_{\delta^{i}r}(x_{0}', t_{0})} |D_{x'}\psi| \leq \sup\limits_{ T_{r} (x_{0}', t_{0}) } |D_{x'}\psi|$ for $i = 0, 1, 2, \cdots$. So from the following computation

we obtain that

From the assumption, $(x_{0}, t_{0})$ is a Lebesgue point of $Du$, which implies that

From the definition of $\nu$, we showed that

So we find that Theorem 1.1 holds.

5.   SOLA (Solutions Obtained by Limit of Approximations)

We explain just a sketch proof for Corollary 1.3 because the proof related to SOLA appears in many papers, say [1, Section 5.2], [17, Section 4.3] and [4, Remark 7]. Suppose that $u_{h} \to u$ in $L^{2} \left(-4R^{2}, 0 \, ; \, W^{1, 1} (K_{2R}^{+}) \right)$ and $L^{\infty} \owns \mu_{h} \rightharpoonup \mu$ in the sense of measures satisfying

for every $Q^{+}_{\rho}(x_{0}, t_{0})$, where $\lfloor Q \rfloor$ denotes the parabolic closure of $Q$. We return to (4.15) in the proof of Theorem 1.1 for the size $r/2$ instead of $r$. For $i \in \{ 0, 1, 2, \cdots \}$, replace $u$ and $\mu$ with $u_{h}$ and $\mu_{h}$, respectively. Then we find that

for $i = 0, 1, 2, \cdots$. By sending $h \to \infty$, we find that

for $i = 0, 1, 2, \cdots$, which implies that

for $i = 0, 1, 2, \cdots$, because $\mu \big|_{\mathbb{R}^{n+1} \setminus Q^{+}_{2R}} = 0$. So Corollary 1.3 follows, because $(x_{0}, t_{0})$ is a Lebesgue point of $Du$.

Author contributions

Ho-Sik Lee : Writing-Original draft preparation; Youchan Kim : Writing-Original draft preparation.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

Y. Kim was supported by the 2023 Research Fund of the University of Seoul. H.-S. Lee was supported by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) through (GRK 2235/2 2021 - 282638148) at Bielefeld University. The authors would like to thank the referee for the careful reading of this manuscript, and offering valuable comments for this manuscript.

Conflict of interest

The authors declare there is no conflict of interest.